Calculating Heats of Reaction

1. Calculating Heats of Reaction • Hess’s Law and Heats of Formation

2. Introduction • If we want to measure the heat of reaction of methane, we just burn a known amount of methane gas in oxygen and measure the heat evolved by using a calorimeter. • But, sometimes we don’t want consume the material when we are interested in the heat of reaction. • What is the heat of formation of a diamond?

3. Hess’s Law • There is a way to measure the heat of reaction of a compound in a way that does not require destroying the compound. • We use Hess’s Law: • If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

4. Hess’s Law • The law is easier to use than it is to say. • For example: • Find the heat of reaction to go from diamond to graphite. • C(diamond) → C(graphite)

5. Hess’s Law • C(diamond) → C(graphite) • We can’t measure this directly, because the process is too slow. • But we do know the following information: • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

6. Hess’s Law • C(diamond) → C(graphite) • If we rearrange the first equation ... • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

7. Hess’s Law • C(diamond) → C(graphite) • If we rearrange the first equation ... • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

8. Hess’s Law • C(diamond) → C(graphite) • ... add them together ... • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ

9. Hess’s Law • C(diamond) → C(graphite) • ... add them together ... • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)

10. Hess’s Law • C(diamond) → C(graphite) • ... and simplify ... • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)

11. Hess’s Law • C(diamond) → C(graphite) • ... and simplify ... • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • CO2(g) + C(diamond) + O2(g) → C(graphite) + CO2(g) + O2(g)

12. Hess’s Law • C(diamond) → C(graphite) • ... and simplify ... • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • C(diamond) → C(graphite)

13. Hess’s Law • C(diamond) → C(graphite) • ... we get the desired equation. • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • C(diamond) → C(graphite)

14. Hess’s Law • C(diamond) → C(graphite) • Now, we add the heats of reaction. • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • C(diamond) → C(graphite)

15. Hess’s Law • C(diamond) → C(graphite) • Now, we add the heats of reaction. • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • C(diamond) → C(graphite) ∆H = +393.5 kJ − 395.4 kJ

16. Hess’s Law • C(diamond) → C(graphite) • Now, we add the heats of reaction. • CO2(g) → C(graphite) + O2(g) ∆H = +393.5 kJ • C(diamond) + O2(g) → CO2(g) ∆H = −395.4 kJ • C(diamond) → C(graphite) ∆H = − 1.9 kJ • This is an exothermic process.

17. Hess’s Law • Let’s try another one. • Find ∆H for the reaction: • C(graphite) + ½ O2(g) → CO(g) • given: • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

18. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • Notice that C(graphite) is on the left of both equations. • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

19. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • This means that we will write the equation with C(graphite) the same way in our final equations. • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

20. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • Notice that CO(g) is on the right above and on the left of below. • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

21. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • This means that we will write the equation with CO(g) the opposite way in our final equations. • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO(g) + ½ O2(g) → CO2(g) ∆H = −283.0 kJ

22. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • This means that we will write the equation with CO(g) the opposite way in our final equations. • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

23. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • Next, we add everything together ... spacer spacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ

24. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • Next, we add everything together ... spacer spacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ • C(graphite) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g)

25. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • ... and simplify ... spacer spacerspacerspacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ • C(graphite) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g) ½ O2(g)

26. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • ... and simplify ... spacer spacerspacerspacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ • C(graphite) + ½ O2(g)→ CO(g)

27. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • ... to get the equation that we want. spacer spacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ • C(graphite) + ½ O2(g)→ CO(g)

28. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • Now, we add the heats of reaction together. spacer spacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ • C(graphite) + ½ O2(g)→ CO(g)

29. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • Now, we add the heats of reaction together. spacer spacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ • C(graphite) + ½ O2(g)→ CO(g) ∆H = −393.5 kJ + 283.0 kJ

30. Hess’s Law • C(graphite) + ½ O2(g) → CO(g) • Now, we add the heats of reaction together. spacer spacerspacer • C(graphite) + O2(g) → CO2(g) ∆H = −393.5 kJ • CO2(g) → CO(g) + ½ O2(g) ∆H = +283.0 kJ • C(graphite) + ½ O2(g)→ CO(g) ∆H = −110.5 kJ • This is an exothermic process.

31. Hess’s Law • Let’s try another one. • Find ∆H for the reaction: • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • given: • PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

32. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Notice that P(s) is on the left of both equations. spacer spacer • PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

33. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • This means that we will write the equation with P(s) the same way in our final equations. • PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

34. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Notice that PCl5(s) is on the right above and on the left of below. • PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

35. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • This means that we will write the equation with PCl5(s) the opposite way in our final equations. • PCl5(s) → PCl3(g) + Cl2(g) ∆H = +87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

36. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • This means that we will write the equation with PCl5(s) the opposite way in our final equations. • PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

37. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Also notice that PCl5(g) is has a coefficient of “2” above and “1” of below. • PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

38. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • This means that we need to multiply the equation below by 2. • PCl3(g) + Cl2(g) → PCl5(s) ∆H = −87.9 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

39. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • This means that we need to multiply the equation below by 2. • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

40. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Next, we add everything together ... spacer spacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ

41. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Next, we add everything together ... spacer spacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ • 2 PCl3(s) + 2 Cl2(g) + 2 P(s) + 3 Cl2(g) → 2 PCl5(s) + 2 PCl3(g)

42. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • ... and simplify ... spacer spacerspacerspacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ • 2 PCl3(s) + 2 Cl2(g) + 2 P(s) + 3 Cl2(g) → 2 PCl5(s) + 2 PCl3(g)

43. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • ... and simplify ... spacer spacerspacerspacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ • 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

44. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • ... to get the equation that we want. spacer spacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ • 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

45. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Now, we add the heats of reaction together. spacer spacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ • 2 P(s) + 5 Cl2(g) → 2 PCl5(s)

46. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Now, we add the heats of reaction together. spacer spacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ − 574 kJ

47. Hess’s Law • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) • Now, we add the heats of reaction together. spacer spacerspacer • 2 PCl3(g) + 2 Cl2(g) → 2 PCl5(s) ∆H = −175.8 kJ • 2 P(s) + 3 Cl2(g) → 2 PCl3(g) ∆H = −574 kJ • 2 P(s) + 5 Cl2(g) → 2 PCl5(s) ∆H = −749.8 kJ = −750. kJ • This is a highly exothermic reaction.

48. Standard Heats of Reaction • Changes in enthalpy (∆H) are usually specified at a set of standard conditions. • Standard temperature = 25°C = 298 K • Standard pressure = 1 atm = 101.3 kPa • The change in enthalpy that accompanies the formation of one mole of a compound from its elements under standard conditions is called the standard heat of formation, ∆Hfo.

49. Standard Heats of Reaction • For example: • H2(g) + ½ O2(g) → H2O(l) ∆Hfo = −285.8 kJ • Ca(s) + ½ O2(g) → CaO(s) ∆Hfo = −635 kJ • C(s) + 2 H2(g) → CH4(g) ∆Hfo = −74.86 kJ • ½ N2(g) + ½ O2(g) → NO(g) ∆Hfo = 90.37 kJ

50. Standard Heats of Reaction • For a reaction occurring at standard conditions, you can calculate the heat of reaction by using the standard heats of formation of the reactants and the products. • ∆Ho = ∆Hfo(products) − ∆Hfo(reactants)